K L2-3 新旷野地带

题目地址；

https://ac.nowcoder.com/acm/contest/5587/K

基本思路：

比较简单的组合问题，我们枚举放个极巨化坑，那么每次我们从行里选行，然后每次多选定了一列，接下来能选择的列数就会减少一，也就是能得到也就是,枚举 然后求个逆元算答案就好了。

参考代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define IO std::ios::sync_with_stdio(false)
#define int long long
#define rep(i, l, r) for (int i = l; i <= r; i++)
#define per(i, l, r) for (int i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
#define INF 0x3f3f3f3f

inline int read() {
  int x = 0, neg = 1; char op = getchar();
  while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
  while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
  return neg * x;
}
inline void print(int x) {
  if (x < 0) { putchar('-'); x = -x; }
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
const int maxn = 1e6 + 10;
int fact[maxn];
inline int qsm(int x,int n,int mod) {
  int res = 1;
  while (n > 0) {
    if (n & 1) res = res * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return res;
}
inline int mod_comb(int n,int k,int mod) {
  if(n == k || k == 0) return 1;
  return fact[n] * qsm(fact[k] * fact[n - k] % mod, mod - 2, mod) % mod;
}
const int mod = 1e9 + 7;
int n,m,k;
signed main() {
  IO;
  fact[0] = 0, fact[1] = 1;
  for (int i = 2; i < maxn; i++) fact[i] = fact[i - 1] * i % mod;
  cin >> n >> m >> k;
  int ans = 0;
  rep(i,1,k){
    if(i > n || i > m) break;
    int tmp = mod_comb(n,i,mod) * fact[m] % mod;
    if( m - i > 0 ) tmp *= qsm(fact[m - i],mod - 2,mod) % mod;
    tmp %= mod;
    ans = (ans + tmp) % mod;
  }
  cout << ans << '\n';
  return 0;
}